Respuesta :

gmany
[tex]\dfrac{4-x}{x^2+5x-6}:\dfrac{x^2-11x+28}{x^2+7x+6}=(*)[/tex]

[tex]4-x=-(x-4)\\\\x^2+5x-6=x^2+6x-x-6=x(x+6)-1(x+6)=(x+6)(x-1)\\\\x^2-11x+28=x^2-7x-4x+28=x(x-7)-4(x-7)\\=(x-7)(x-4)\\\\x^2+7x+6=x^2+6x+x+6=x(x+6)+1(x+6)=(x+6)(x+1)\\\\(*)=\dfrac{-(x-4)}{(x+6)(x-1)}:\dfrac{(x-7)(x-4)}{(x+6)(x+1)}[/tex]
[tex]=\dfrac{-(x-4)}{(x+6)(x-1)}\cdot\dfrac{(x+6)(x+1)}{(x-7)(x-4)}=\dfrac{-1}{x-1}\cdot\dfrac{x+1}{x-7}\\\\=-\dfrac{x+1}{x^2-7x-x+7}=-\dfrac{x+1}{x^2-8x+7}[/tex]

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